On Sat, Feb 04, 2017 at 01:50:12PM -0600, Luis Mochan wrote:
> On Sat, Feb 04, 2017 at 02:17:44PM -0500, Chris Marshall wrote:
> > I think $P(a) != $P(b) will work.
It did work, in a way.

What I now don't understand is the use of inplace. My routine has two
outputs (one of them an integer pdl). If I now call it as
   ($y, $i)=f($x);
it recognizes that $x is a different pdl than the output and
copies the data before calling the fortran routine.

I can also call the routine as
   f($x, $y=PDL->null, $i=PDL->null);
with the same result.

If I call it as
   $y=$x; f($y,$y, $i=PDL->null);
it recognizes that the expected output $y and the input are the same pdl and
calls the fortran routine without copying, producing the correct
result.

However, if I call it as
   $y=$x; $i=f($y->inplace);
I get a segmentation fault, and if I call it as 
   $y=$x; f($y->inplace, $i->PDL->null);
I get a remainder the f expects three arguments, not two.

I did have in my pp_def the line Inplace0>['x','y'].

Is it that inplace is designed for one output routines only, or am I
making something wrong?

For the time being, I guess I'll pass only the input arguments
(without using inplace) or all input and output arguments.

Best regards,
Luis





> > 
> > It would probably work with $PDL(a) == $PDL(b) as well but I don't know.
> > 
> > A quick test should answer the question.  I'm hoping Craig or
> > another PDL::PP expert can weigh in from their experience and
> > expertise.
> Thanks, I'll test these solutions.
> Regards,
> Luis
> 
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